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fasiddique
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| for (uint64_t i = 0; i < input.size(); i++) { | ||
| hostoutput[i + 1] = hostoutput[i] + input[i + 1]; |
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This is an out-of-bounds access. Depending on the compiler and operating system, it may either get segmentation fault, or produce error-prone results.
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prefix-sum.out(9981,0x1f217cf40) malloc: Incorrect checksum for freed object 0x12200c800: probably modified after being freed.
Corrupt value: 0xffc00
prefix-sum.out(9981,0x1f217cf40) malloc: *** set a breakpoint in malloc_error_break to debug
Abort trap: 6
fasiddique
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May 19, 2025
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| for (uint64_t i = 0; i < intermeadiate_results.size(); ++i) { | ||
| if (i % 2 == 0) | ||
| even.push_back(intermeadiate_results[i]); | ||
| else | ||
| odd.push_back(intermeadiate_results[i]); | ||
| } |
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It's recommended to avoid using push_back for efficiency purpose, because this allocates memory dynamically. Since the vector size is known beforehand, we can use reside and indexing.
fasiddique
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May 19, 2025
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| intermeadiate_results.push_back(input[i]); | ||
| host_device_merged.push_back(input[i]); |
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Why do we need to have three copies of the input vector?
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Please the updated version.